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Alinara [238K]
3 years ago
16

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

force is 3.5 × 10–2 N, how fast is the charge moving?
Physics
1 answer:
larisa86 [58]3 years ago
4 0

Answer:

The charge is moving with the  velocity of 1.1\times10^{4}\ m/s.

Explanation:

Given that,

Charge q =8.4\times10^{-4}\ C

Angle = 35°

Magnetic field strength B=6.7\times10^{-3}\ T

Magnetic force F=3.5\times10^{-2}\ N

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

F = qvB\sin\theta

v = \dfrac{F}{qB\sin\theta}

Where,

F =  Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}

v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}

v = 10910.36\ m/s

v = 1.1\times10^{4}\ m/s

Hence, The charge is moving with the  velocity of 1.1\times10^{4}\ m/s.

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Explanation:

hope this helps

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